Find the work done by winding up a hanging cable of length 100 ft and weight density 5 lb/ft.

To find the work done by winding up a hanging cable, we need to calculate the work done in lifting each infinitesimal segment of the cable from its initial height to the top.

### Key information:
- Length of the cable, \( L = 100 \) feet.
- Weight density of the cable, \( \rho = 5 \) lb/ft.

#### Steps to calculate the work done:

1. **Consider an infinitesimal segment**:  
   Let \( x \) be the distance from the bottom of the cable to the point where we are winding the cable.  
   The weight of an infinitesimal segment of the cable of length \( dx \) is:  
   \[
   dW = \rho \, dx = 5 \, dx \text{ lb.}
   \]

2. **Height to which the segment is lifted**:  
   The height that each infinitesimal segment is lifted depends on its position. A segment at distance \( x \) from the bottom must be lifted a distance of \( L - x \), or \( 100 - x \) feet.

3. **Infinitesimal work**:  
   The infinitesimal work required to lift this segment is the weight of the segment multiplied by the distance it is lifted:
   \[
   dW = 5 \, (100 - x) \, dx.
   \]

4. **Total work**:  
   To find the total work done, integrate from \( x = 0 \) (bottom of the cable) to \( x = 100 \) (top of the cable):
   \[
   W = \int_0^{100} 5 \, (100 - x) \, dx.
   \]

5. **Solve the integral**:
   \[
   W = 5 \int_0^{100} (100 - x) \, dx
   \]
   \[
   W = 5 \left[ 100x - \frac{x^2}{2} \right]_0^{100}.
   \]

6. **Evaluate the expression**:
   \[
   W = 5 \left[ (100 \times 100 - \frac{100^2}{2}) - (100 \times 0 - \frac{0^2}{2}) \right]
   \]
   \[
   W = 5 \left[ 10000 - 5000 \right]
   \]
   \[
   W = 5 \times 5000 = 25000 \text{ foot-pounds}.
   \]

### Final Answer:
The total work done in winding up the hanging cable is \( \boxed{25,000} \) foot-pounds.

Would you like further clarification on any step or more details on the concept?

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**Related Questions**:
1. How would the work change if the weight density of the cable was doubled?
2. What would the work be if only half the cable (50 feet) was wound up?
3. How is the work formula affected if the cable has varying density along its length?
4. Can you explain how potential energy relates to the work in this case?
5. How does the formula change if the cable is wound up on an inclined plane instead of vertically?

**Tip**: When dealing with work in physics, always pay attention to how force and distance vary along the path of the object being moved!

Find the work done by winding up a hanging cable of length 100 ft and weight density 5 lb/ft using integration. This problem involves key calculus concepts such as infinitesimal work and definite integrals. Suitable for students in Calculus I.

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